3 Answers
3

The photon itself does not have any physical extent: it just travels along a line, as a point particle so far as we can tell.

It carries with it two fields: the electric field $\vec E$ and the magnetic field $\vec B$. These are both vector fields, in that they have vector values at every point in space. Off the trajectory of the photon, their value is the zero vector. At every point along the trajectory of the photon, however, you can associate a vector value for both the electric and the magnetic field. As a concrete example, take a linearly polarized photon traveling along the z-axis. The electric field oscillates in the direction of the x-axis, and the magnetic field oscillates in the direction of the y-axis. So light is often depicted in textbooks as having two oscillations around the axis of the trajectory, representing the electric and magnetic fields. The fields themselves are only present along the trajectory of the photon, though.

EDIT: The above description is for a single photon, or for an infinitely narrow beam of light. In practice, that doesn't exist. In practice (for example, a laser), you have a beam of light with some finite extent. The actual intensity of photons might vary with position across the beam of light. So in most experimental setups, there will be some spatial extent to your beam of light.

I believe the other current answer is wrong on several levels. Consider the cross section limit in Mie (spherical particle) scattering divided by the geometric cross section of the sphere. The limiting value is 2 for large spheres. What this means is that photons that don't 'hit' the sphere are scattered by the sphere. Since Mie scattering is easily observed, this implies a horizontal extent for a photon. It does NOT mean that the cross section is measured from two directions. It is possible for resonances to reach values that are very very high - much more than 2. Resonances are observable 8 x the geometric cross section and can be calculated exactly with Mie scattering programs.

Another argument that photons do not propagate as point particles is the resolution of a telescope. If one grinds the mirror in a telescope to an infinitely perfect shape so that all photons are theoretically geometrically focused to an infintesimal point, why does this not work which is easily seen with the eye? The answer is that the edges of the telescope are 'seen' by the single photons whether it is 6 inches in diameter or it is 18 feet in diameter. The angular resolution directly proportional to the wavelength of light and inversely proportional to the Diameter of the Mirror (lens).

Or to put it more simply, Merzbacher once told me directly that photons have an infinite horizontal extent.

These simply observable experimental points argue strongly that Merzbacher was right. Furthermore, it illustrates the wave-particle duality. Photons are not 'just' point particles propagating on a line, and this can be easily demonstrated experimentally.

Huygens principle - Every point on a wave-front may be considered a source of secondary spherical wavelets which spread out in the forward direction at the speed of light. The new wave-front is the tangential surface to all of these secondary wavelets.

This is true even for single photons, is consistent with the examples, and to ignore it leads to endless paradoxes. So if a wave front is terminated, say at the edges of a telescope, or by a diffraction edge, then the photon is deviated.

have you not heard of the wave particle duality? It holds for photons too, it has a wave nature that appears in the two slit experiment even with singe photons, and a particle nature that appears in the photoelectric effect.
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anna vNov 6 '14 at 17:26

Elementary particles are quantum mechanical entities, not classical particles or waves. Quantum mechanical entities have dual manifestations, either as point particles at specific (x,y,z) in some observations, or as probability waves: the probability distribution of finding them at a specific (x,y,z) in some experiments has a wave like manifestation with interferences etc. The double slit experiment is an example of this dual nature. The points on the screen ( seet the image in the paragraph on interference of individual particles) are the footprint in x,y of an electron, the accumulated distribution shows the wave nature of the probability of finding it at x,y . All these effects are calculable and predicted using quantum mechanical equations.

Photons due to their having zero mass, have the attribute that their energy is E=h*nu where nu is the frequency displayed by the light as measured macroscopically . They also have spin 1, either +1 or -1 projections (no 0). Light obeys the classical maxwell's equations and is composed by zillions of photons. Experiments have been done reducing light intensity to a single photon at a time so there is no doubt about this statement.

The connection between the classical electromagnetic equations which have polarized light in space and varying electric and magnetic fields, and a single photon come from the quantum mechanical wavefunction of the photon, which is given by the Maxwell equations turned into quantum mechanical ones by the operator formalism, acting on the wave function of the photon. This is done in the format of the electromagnetic potential. This gives to the photon the energy E=h*nu and a phase to characterize it. These phases will produce the polarization observed in space for the classical electromagnetic wave. How the classical wave emerges from the quantum mechanical particle ensemble is sketched in this blog entry.

If light (a photon) is linearly polarized, say vertically, does it have some vertical spatial extent (perhaps in amplitude)?

The simple answer is that no, the photon as an elementary particle does not have a spatial extent, it is a point particle moving with the velocity of light. The spatial extent observed macroscopically as classical polarization is an emergent phenomenon from the coherent confluence of a large ensemble of photons.